Tree2cycle

Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1730 Accepted Submission(s): 401


Problem Description A tree with N nodes and N-1 edges is given. To connect or disconnect one edge, we need 1 unit of cost respectively. The nodes are labeled from 1 to N. Your job is to transform the tree to a cycle(without superfluous edges) using minimal cost.

A cycle of n nodes is defined as follows: (1)a graph with n nodes and n edges (2)the degree of every node is 2 (3) each node can reach every other node with these N edges.
Input The first line contains the number of test cases T( T<=10 ). Following lines are the scenarios of each test case.
In the first line of each test case, there is a single integer N( 3<=N<=1000000 ) - the number of nodes in the tree. The following N-1 lines describe the N-1 edges of the tree. Each line has a pair of integer U, V ( 1<=U,V<=N ), describing a bidirectional edge (U, V).

Output For each test case, please output one integer representing minimal cost to transform the tree to a cycle.

Sample Input

1
4
1 2
2 3
2 4

Sample Output

3
Hint
In the sample above, you can disconnect (2,4) and then connect (1, 4) and
(3, 4), and the total cost is 3.
 

Source 2013 ACM/ICPC Asia Regional Online —— Warmup
Recommend liuyiding | We have carefully selected several similar problems for you: 5368 5367 5366 5365 5364 具體思路http://blog.csdn.net/dongdongzhang_/article/details/11395305 額，，不能算dp吧，充其量算個樹上的亂搞。。會爆棧。。開棧就過了 ac代碼

#include
#include
#pragma comment(linker, /STACK:102400000,102400000)
int head[1000010],vis[1000010],cnt,ans;
struct s
{
	int u,v,next;
}edge[1000010<<1];
void add(int u,int v)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
int dfs(int u)
{
	vis[u]=1;
	int sum=0;
	for(int i=head[u];i!=-1;i=edge[i].next)
	{
		int v=edge[i].v;
		if(!vis[v])
		{
			sum+=dfs(v);
		}
	}
	if(sum>=2)
	{
		if(u==1)
			ans+=(sum-2)*2;
		else
			ans+=(sum-1)*2;
		return 0;
	}
	return 1;
}
int main()
{
	int t;
	scanf(%d,&t);
	while(t--)
	{
		int n;
		cnt=0;
		ans=0;
		memset(head,-1,sizeof(head));
		memset(vis,0,sizeof(vis));
		scanf(%d,&n);
		int i;
		for(i=1;i